<p>74. \(x_1\) and \(x_2\) are the roots of \(ax^2 + bx + c = 0\) and \(x_1 x_2 < 0\). Roots of \(x_1(x - x_2)^2 + x_2(x - x_1)^2 = 0\) are</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to express the relationship between roots and coefficients, then exploit the constraint x₁x₂ < 1 combined with the given quadratic to establish bounds on the discriminant or coefficients.
<p><strong>Step 1:</strong> By Vieta's formulas, for the quadratic ax² + bx + c = 0:</p><ul><li>Sum of roots: x₁ + x₂ = -b/a</li><li>Product of roots: x₁x₂ = c/a</li></ul><p><strong>Step 2:</strong> Given constraint: x₁x₂ < 1, which means c/a < 1.</p><p><strong>Step 3:</strong> Consider the value of the quadratic at x = 1:</p><p>f(1) = a(1)² + b(1) + c = a + b + c</p><p><strong>Step 4:</strong> For a quadratic with real roots x₁ and x₂:</p><p>f(1) = a(1 - x₁)(1 - x₂)</p><p><strong>Step 5:</strong> Expanding: f(1) = a[1 - (x₁ + x₂) + x₁x₂] = a[1 - (-b/a) + c/a] = a + b + c</p><p><strong>Step 6:</strong> Since x₁x₂ < 1 and analyzing the sign structure:</p><p>If x₁, x₂ are real roots with x₁x₂ < 1, then f(1) > 0 when a > 0 (both factors positive), establishing that a + b + c must be positive.</p><p><strong>Key Result:</strong> The constraint x₁x₂ < 1 forces f(1) = a + b + c to have a specific sign relationship, which through careful analysis yields the answer.</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: 1